Integrand size = 86, antiderivative size = 24 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {9 e^{5-x} (2+\log (4))}{(5+x)^2 \log \left (x^2\right )} \]
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {9 e^{5-x} (2+\log (4))}{(5+x)^2 \log \left (x^2\right )} \]
Integrate[(E^5*(-180 - 36*x) + E^5*(-90 - 18*x)*Log[4] + (E^5*(-126*x - 18 *x^2) + E^5*(-63*x - 9*x^2)*Log[4])*Log[x^2])/(E^x*(125*x + 75*x^2 + 15*x^ 3 + x^4)*Log[x^2]^2),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-x} \left (\left (e^5 \left (-18 x^2-126 x\right )+e^5 \left (-9 x^2-63 x\right ) \log (4)\right ) \log \left (x^2\right )+e^5 (-36 x-180)+e^5 (-18 x-90) \log (4)\right )}{\left (x^4+15 x^3+75 x^2+125 x\right ) \log ^2\left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (\left (e^5 \left (-18 x^2-126 x\right )+e^5 \left (-9 x^2-63 x\right ) \log (4)\right ) \log \left (x^2\right )+e^5 (-36 x-180)+e^5 (-18 x-90) \log (4)\right )}{x \left (x^3+15 x^2+75 x+125\right ) \log ^2\left (x^2\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-x} \left (\left (e^5 \left (-18 x^2-126 x\right )+e^5 \left (-9 x^2-63 x\right ) \log (4)\right ) \log \left (x^2\right )+e^5 (-36 x-180)+e^5 (-18 x-90) \log (4)\right )}{x (x+5)^3 \log ^2\left (x^2\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {9 e^{5-x} (2+\log (4)) \left (-x (x+7) \log \left (x^2\right )-2 (x+5)\right )}{x (x+5)^3 \log ^2\left (x^2\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 9 (2+\log (4)) \int -\frac {e^{5-x} \left (2 (x+5)+x (x+7) \log \left (x^2\right )\right )}{x (x+5)^3 \log ^2\left (x^2\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -9 (2+\log (4)) \int \frac {e^{5-x} \left (2 (x+5)+x (x+7) \log \left (x^2\right )\right )}{x (x+5)^3 \log ^2\left (x^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -9 (2+\log (4)) \int \left (\frac {e^{5-x} (x+7)}{(x+5)^3 \log \left (x^2\right )}+\frac {2 e^{5-x}}{x (x+5)^2 \log ^2\left (x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -9 (2+\log (4)) \left (\frac {2}{25} \int \frac {e^{5-x}}{x \log ^2\left (x^2\right )}dx-\frac {2}{5} \int \frac {e^{5-x}}{(x+5)^2 \log ^2\left (x^2\right )}dx-\frac {2}{25} \int \frac {e^{5-x}}{(x+5) \log ^2\left (x^2\right )}dx+2 \int \frac {e^{5-x}}{(x+5)^3 \log \left (x^2\right )}dx+\int \frac {e^{5-x}}{(x+5)^2 \log \left (x^2\right )}dx\right )\) |
Int[(E^5*(-180 - 36*x) + E^5*(-90 - 18*x)*Log[4] + (E^5*(-126*x - 18*x^2) + E^5*(-63*x - 9*x^2)*Log[4])*Log[x^2])/(E^x*(125*x + 75*x^2 + 15*x^3 + x^ 4)*Log[x^2]^2),x]
3.17.36.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 7.97 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42
method | result | size |
parallelrisch | \(\frac {\left (36 \,{\mathrm e}^{5} \ln \left (2\right )+36 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{2 \ln \left (x^{2}\right ) \left (x^{2}+10 x +25\right )}\) | \(34\) |
risch | \(\frac {36 i \left (1+\ln \left (2\right )\right ) {\mathrm e}^{5-x}}{\left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )\right ) \left (x^{2}+10 x +25\right )}\) | \(76\) |
int(((2*(-9*x^2-63*x)*exp(5)*ln(2)+(-18*x^2-126*x)*exp(5))*ln(x^2)+2*(-18* x-90)*exp(5)*ln(2)+(-36*x-180)*exp(5))/(x^4+15*x^3+75*x^2+125*x)/exp(x)/ln (x^2)^2,x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {18 \, {\left (e^{5} \log \left (2\right ) + e^{5}\right )} e^{\left (-x\right )}}{{\left (x^{2} + 10 \, x + 25\right )} \log \left (x^{2}\right )} \]
integrate(((2*(-9*x^2-63*x)*exp(5)*log(2)+(-18*x^2-126*x)*exp(5))*log(x^2) +2*(-18*x-90)*exp(5)*log(2)+(-36*x-180)*exp(5))/(x^4+15*x^3+75*x^2+125*x)/ exp(x)/log(x^2)^2,x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {\left (18 e^{5} \log {\left (2 \right )} + 18 e^{5}\right ) e^{- x}}{x^{2} \log {\left (x^{2} \right )} + 10 x \log {\left (x^{2} \right )} + 25 \log {\left (x^{2} \right )}} \]
integrate(((2*(-9*x**2-63*x)*exp(5)*ln(2)+(-18*x**2-126*x)*exp(5))*ln(x**2 )+2*(-18*x-90)*exp(5)*ln(2)+(-36*x-180)*exp(5))/(x**4+15*x**3+75*x**2+125* x)/exp(x)/ln(x**2)**2,x)
Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {9 \, {\left (\log \left (2\right ) + 1\right )} e^{\left (-x + 5\right )}}{{\left (x^{2} + 10 \, x + 25\right )} \log \left (x\right )} \]
integrate(((2*(-9*x^2-63*x)*exp(5)*log(2)+(-18*x^2-126*x)*exp(5))*log(x^2) +2*(-18*x-90)*exp(5)*log(2)+(-36*x-180)*exp(5))/(x^4+15*x^3+75*x^2+125*x)/ exp(x)/log(x^2)^2,x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {18 \, {\left (e^{\left (-x + 5\right )} \log \left (2\right ) + e^{\left (-x + 5\right )}\right )}}{x^{2} \log \left (x^{2}\right ) + 10 \, x \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )} \]
integrate(((2*(-9*x^2-63*x)*exp(5)*log(2)+(-18*x^2-126*x)*exp(5))*log(x^2) +2*(-18*x-90)*exp(5)*log(2)+(-36*x-180)*exp(5))/(x^4+15*x^3+75*x^2+125*x)/ exp(x)/log(x^2)^2,x, algorithm=\
Time = 14.96 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.62 \[ \int \frac {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx=\frac {9\,\left (\ln \left (2\right )+1\right )\,\left (7\,x\,{\mathrm {e}}^{5-x}+x^2\,{\mathrm {e}}^{5-x}-7\,x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5-x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\right )}{{\left (x+5\right )}^3}+\frac {9\,\left (\ln \left (2\right )+1\right )\,\left (10\,{\mathrm {e}}^{5-x}+2\,x\,{\mathrm {e}}^{5-x}\right )}{\ln \left (x^2\right )\,{\left (x+5\right )}^3} \]
int(-(exp(-x)*(log(x^2)*(exp(5)*(126*x + 18*x^2) + 2*exp(5)*log(2)*(63*x + 9*x^2)) + exp(5)*(36*x + 180) + 2*exp(5)*log(2)*(18*x + 90)))/(log(x^2)^2 *(125*x + 75*x^2 + 15*x^3 + x^4)),x)